Buzz Bloom said:
$$dt = \frac {da} {H_0 a \sqrt {M/a3+L}} = \frac {{\sqrt {a}} da} {H_0 \sqrt {M+L a^3}}$$
I make the following substitutions.
##{x = a^3,}## ##{a = x^{1/3}},## ##{da = dx / 3x^{2/3}},##
(I am not going to use ##Q##.) The substitution that I prefer for this is
$$\frac{L}{M} a^3 = \sinh^2 u .$$
I make this substitution in order to make use of the identity ##\cosh^2 u - \sinh^2 u = 1##. Differentiate:
$$\begin{align}
3\frac{L}{M} a^2 da &= 2\sinh u \frac{d}{du} \sinh u du \\
&= 2\sinh u \cosh u du \\
da &= \frac{2M}{3L} a^{-2} \sinh u \cosh u du
\end{align}$$
Substituting these into the integral gives
$$\begin{align}
\frac {{\sqrt {a}} da} {H_0 \sqrt {M + L a^3}} &= \frac {{\sqrt {a}} da} {H_0 \sqrt {M} \sqrt{1 + \frac{L}{M} a^3}} \\
&= \frac {{\sqrt {a}} \frac{2M}{3L} a^{-2} \sinh u \cosh u du} {H_0 \sqrt {M} \sqrt{1 + \sinh^2 u}} .
\end{align}$$
Now (if you want to proceed with this method), combine the ##a## s, and then replace them by something in terms of ##u## using my original substitution. Also, use the (hyperbolic) trig identity that I gave.
This should result in an expression for ##t\left(a\right)## that involves an inverse ##\sinh##, and that, if desired, can reexpressed using ##\ln## s.